The equation of a circle $C$ is $x^2+y^2-18x-6y+89 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-18x) + (y^2-6y) = -89$ $(x^2-18x+81) + (y^2-6y+9) = -89 + 81 + 9$ $(x-9)^{2} + (y-3)^{2} = 1 = 1^2$ Thus, $(h, k) = (9, 3)$ and $r = 1$.